VI. Suppose 5% of the population have a certain disease. A laboratory blood test gives a positive reading for 95% of people who have the disease and 10% positive reading of people who do not have the disease. a. What is the probability of testing positive? b. What is the probability that a randomly selected person does not has the disease given that this person is testing positive? Solution PROBABILITY OF GETTING DISEASE ,P(A)=0.05 PROBABILITY OF NOT GETTING DISEASE,P(B)=(1-0.05)=0.95 LET E BE EVENT OF OCCURING POSITIVE READING PROBABILITY OF GETTING POSITIVE READING who have the disease ,P(E/A)=0.95 PROBABILITY OF GETTING POSITIVE READING who DO NOT have the disease,P(E/B)=0.10 A)the probability of testing positive,P(E) B)the probability that a randomly selected person does not has the disease given that this person is testing positive,P(B/E) FROM BAYES THEOREM P(B/E)=P(B)*P(E/B)/[P(B)*P(E/B)+P(A)*P(E/A)] =[0.95*0.10]/[(0.95*0.10)+(0.05*0.95)] =[0.095]/[0.095+0.0475] =0.667 HENCE,P(B/E)=0.67.